LCAO-MO Correlation Diagrams

Transcription

1 LCAO-MO Correlation Diagrams (Linear Combination of Atomic Orbitals to yield Molecular Orbitals) For (Second Row) Homonuclear Diatomic Molecules (X 2 ) - the following LCAO-MO s are generated: LCAO MO symbol 1s A 1s B σ 1s 1s A - 1s B σ * 1s 2s A 2s B σ 2s 2s A - 2s B σ * 2s 2p x,a 2p x,b π 2p 2p x,a - 2p x,b π * 2p 2p y,a 2p y,b π 2p 2p y,a - 2p y,b π * 2p 2p z,a 2p z,b σ 2p 2p z,a - 2p z,b σ * 2p For the above LCAO-MO combinations, the coordinate system is chosen such that the z axis is along the horizontal direction and is considered the internuclear ( bond ) axis. Here, the y-axis is considered to be along the vertical direction and the x-axis is considered to be perpendicular to the plane of the page. This choice is, of course, arbitrary. The correlation diagrams showing the energy ordering and relationship between the Atomic Orbitals (AO s) and resultant Molecular Orbitals (MO s) for several situations are listed below. Each MO can maximally contain two (2) electrons - with opposite spins ( ) - as required by the Pauli Exclusion Principle (PEP). Also, degenerate MO s - when occupied - will follow Hund s Rule in order to achieve a ground state (energy-preferred) electron configuration. There are two schemes - I and II below. Scheme I applies for Li through N (and their ions), inclusive and Scheme II applies for O through Ne (and their ions), inclusive. As can be seen, the difference lies in the relative energy ordering of the π 2p MO s versus the σ 2p MO s. For Z 7 atoms, the (degenerate) π 2p MO s are lower in energy than σ 2p MO. For Z 8 atoms, the (degenerate) π 2p MO s are greater in energy than σ 2p MO. The reason has to do with energy stabilization by attenuation of electron-electron repulsion. In Scheme I, the π 2p MO s are filled before the σ 2p MO because the electron density in the π 2p MO s are concentrated (between the atoms) away from (i.e., above and below) the internuclear axis. This leads to a reduction in the electron-electron repulsions. This is particularly important since the electrons in the already - occupied σ 2s - σ 2s * MO s will interact less strongly with electrons in the π2p MO s than those in the σ 2p MO s (electron density also directed along the internuclear axis). In Scheme II - followed by atoms toward the end of the second row - the already occupied σ 2s - σ 2s * MO s are drawn closer ( tighter ) due to the greater nuclear charge (Z). For Z 8, this is enough so that the σ 2p MO s will interact less strongly with the σ 2s - σ 2s * MO s. Hence, the σ 2p MO will be lower in energy than the π 2p MO s. [After discussing Scheme II, your text sometimes follows Scheme I - for simplicity - for all second row diatomics (and their ions) in some of the homework problems.] By paying attention to the PEP and Hund s Rule - as mentioned above - we fill the MO s from bottom - up in an Aufbau manner for our chosen MO scheme. This will give us ground state (MO) electron configurations. 1

6 Where are the bonds?? In order to determine how many bonds will form between the atoms of the diatomic molecule (in the spirit of the Valence Bond - localized electron pair bond model), we define the parameter BOND ORDER: BOND ORDER = B.O. ({# of e s in Bonding MO s} {# of e s in Antibonding ( * ) MO s}) 2 Prove for yourself that the Bond Order for N 2, with a ground state valence shell electron configuration of: σ 2s 2 (σ 2s * ) 2 π 2p 4 σ 2p 2 is 3.0, i.e., a triple bond. Note, this is exactly what the Valence Bond model would also predict! If a molecule has a bond order of ZERO (0), then M.O. theory is telling us that the MOLECULE WILL NOT FORM (i.e., no bonds will be created, since the number of electrons in antibonding M.O. s exactly cancels the number of electrons in bonding M.O. s). For example: He 2 and Be 2 have B.O. = 0, so these molecules will not form, i.e., they are not energetically stable. Finally, if a molecule has an odd number of electrons, such as CN (a radical), then we can have a fractional bond order. Prove for yourself that CN, with M.O. ground state valence shell configuration: σ 2s 2 (σ 2s * ) 2 π 2p 4 σ 2p 1, has a bond order of 2.5. Bond orders can be used to predict the relative STRENGTH and relative LENGTHS of BONDS. The relationship is: THE GREATER THE BOND ORDER, THE SHORTER AND STRONGER THE BOND. Hence, B 2, C 2, and N 2,which have bond orders of 1.0, 2.0, and 3.0, respectively, have BOND LENGTHS of 159 pm, 124 pm, and 110 pm, respectively. The BOND ENERGIES of B 2, C 2, and N 2 are: 289 kj/mol, 599 kj/mol, and 941 kj/mol, respectively. Paramagnetism & Diamagnetism: It is important to realize that M.O. theory allows a molecule with an EVEN (as well as an odd) number of electrons to have unpaired electrons. According to the Valence Bond model all molecules with an EVEN number of electrons will have the electrons either PAIRED in bonds or in lone PAIRS - and are predicted to be DIAMAGNETIC. Thus, Valence Bond Theory predicts that O 2 (with 16 total electrons or 12 valence electrons) will have a double bond plus two (2) lone pairs around each Oxygen atom. According to M.O. theory, the ground state valence shell electron configuration is: σ 2s 2 (σ 2s * ) 2 σ 2p 2 π 2p 4 (π 2p * ) 2 - where each of the two (2) electrons in the π 2p * M.O. are UNPAIRED - according to Hund s Rule. Hence, Valence Bond theory would predict that the oxygen molecule is DIAMAGNETIC (no unpaired electrons); whereas Molecular Orbital theory would predict that the oxygen molecule should be PARAMAGNETIC (two (2) unpaired electrons). Experiment tells us that O 2 is attracted by a magnetic field, i.e., it is PARAMAGNETIC. Thus, in this case, M.O. theory better rationalizes the bonding picture in O 2. Note that both V.B. theory and M.O. theory predict that the oxygen atoms are attached by a double bond - check this for yourself by drawing the Lewis structure and also by calculating the bond order from the ground state M.O. electron configuration (listed above). 6

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